songtian19800919
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共回答了19个问题采纳率:89.5% 举报
设z=x+yi,
(1+3i)(x+yi)=(x-3y)+(3x+y)i
为纯虚数,所以x-3y=0,x=3y,
w=(x+yi)/(2+i)
|w|^2=(x^2+y^2)/(2^2+1^2)=10y^2/5=2y^2
5√2=|w|=√2|y|,
|y|=5,
y=±5
x=±15
w=±5(3+i)/(1+2i)
=±5(3+i)(1-2i)/[(1+2i)(1-2i)]
=±5(5-5i)/5
=±(5-5i)
w=±(5-5i).
1年前
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